Relay-logic controlled printing machine, 110 vac.

A 300mfd (150v) electrolytic cap is used together with 25K (25 W) ohm
wirewound pot and a 5K ohm relay coil to control a variable time delay before
the relay opens.

The cap is shot (measures less than 1mfd). I'm having a hard time finding a
replacement locally. (Yeah, I know: Digi-Key, et al) But I need to get this
up and running today.

What other common types can I use in place of this can-type cap? (It has 2
wings that rivet to the sides of the opening in the chassis and solder terms
make the connection below the chassis.) Maybe poly-somethings, two in series
or parallel?

It's a Mallory FP119A.

Can I use a non-polar electro? (The wiring diagram shows a polarity
indicator.)

Or a motor-start cap? ::

http://tinyurl.com/y9d3ks6

Thanks,
Dave

On Tue, 06 Oct 2009 22:42:57 -0700, DaveC wrote:

Relay-logic controlled printing machine, 110 vac.

A 300mfd (150v) electrolytic cap is used together with 25K (25 W) ohm
wirewound pot and a 5K ohm relay coil to control a variable time delay
before the relay opens.

The cap is shot (measures less than 1mfd). I'm having a hard time
finding a replacement locally. (Yeah, I know: Digi-Key, et al) But I
need to get this up and running today.

What other common types can I use in place of this can-type cap? (It has
2 wings that rivet to the sides of the opening in the chassis and solder
terms make the connection below the chassis.) Maybe poly-somethings, two
in series or parallel?

It's a Mallory FP119A.

Can I use a non-polar electro? (The wiring diagram shows a polarity
indicator.)

Or a motor-start cap? ::

http://tinyurl.com/y9d3ks6

Thanks,
Dave

On the face of it anything that'll get up to 330mfd and withstand 150V
should do. That's a honking big cap, and anything that's not a 'lytic
will probably be bigger.

--
www.wescottdesign.com

DaveC wrote in
:

Relay-logic controlled printing machine, 110 vac.

A 300mfd (150v) electrolytic cap is used together with 25K (25 W) ohm
wirewound pot and a 5K ohm relay coil to control a variable time delay
before the relay opens.

The cap is shot (measures less than 1mfd). I'm having a hard time
finding a replacement locally. (Yeah, I know: Digi-Key, et al) But I
need to get this up and running today.

What other common types can I use in place of this can-type cap? (It has
2 wings that rivet to the sides of the opening in the chassis and solder
terms make the connection below the chassis.) Maybe poly-somethings, two
in series or parallel?

It's a Mallory FP119A.

Can I use a non-polar electro? (The wiring diagram shows a polarity
indicator.)

Or a motor-start cap? ::

http://tinyurl.com/y9d3ks6


A motor start cap should be ok. They're meant for intermittent use under
heavy load so they don't accept heavy ripple currents for lomg, but this one
won't have to, with a few tens of kilohms between it and the mains.

"DaveC" wrote in message
...
Relay-logic controlled printing machine, 110 vac.

A 300mfd (150v) electrolytic cap is used together with 25K (25 W) ohm
wirewound pot and a 5K ohm relay coil to control a variable time delay
before
the relay opens.

The cap is shot (measures less than 1mfd). I'm having a hard time finding
a
replacement locally. (Yeah, I know: Digi-Key, et al) But I need to get
this
up and running today.

What other common types can I use in place of this can-type cap? (It has 2
wings that rivet to the sides of the opening in the chassis and solder
terms
make the connection below the chassis.) Maybe poly-somethings, two in
series
or parallel?

It's a Mallory FP119A.

Can I use a non-polar electro? (The wiring diagram shows a polarity
indicator.)

Or a motor-start cap? ::

http://tinyurl.com/y9d3ks6


You can potentially vary the values of the resistor and capacitor greatly
depending on how much drive is needed. As long as the time constant R*C is
the approximately the same it should approximately work. Obviously if it's
driving a relay then it needs enough current to turn it on. So your probably
not going to be able to vary R much unless it wasn't chosen well in the
first place.

The electrolytic caps were used because of the high voltage and high
capacitance. I doubt you can find anything but an electrolytic to replace it
with. Remember you can put them in parallel to increase the capacitance.
Just get you two, three, or how ever many you need to get approximately
300uF at 150V. I woudln't get any lower voltage and try to use them in
series because it is usually more trouble than it's worth.

Note, you can replace this simple delay circuit a mosfet who's gate is
charged up by a lot smaller capacitor and larger resistor.

Basically the gate is turned on in a similar time frame as the original but
requires virtually no current to do so. The mosfet then turns on which turns
on the relay.

All you need is a resistor(actually two), a cap, a suitable mosfet that can
handle the current and voltage(relatively easy to find for a few dollars),
and a diode for a snubber.

Since your input is 150V or so you'll have to reduce it for the mosfet's
gate voltage. I'll want to use a P ch but could probably put an Nch on the
low side of the coil.

R1
|
+-----+---mosfet Gate
| |
| C
R2 gnd
|
gnd

(use a fixed width font to view)

The R1 and R2 form a voltage divider. You'll want something about 12 to 1 so
that the mosfets gate only sees at most about 15V. A large voltage spike
could ruin the mosfet gate so technically it would be best to proect it. You
can do this by adding a 15-20V zener from the gate to ground.

Add the diode across the mosfet drain to source(look up diode snubber). Most
power mosfets actually have a diode built in to do this though.

In any case if you want to go this route then I'm sure someone can whip you
up a circuit. It's relatively simple and should do the trick better than the
large cap(And probably cheaper). Theres just a few small issues such as
getting hte polarity right and getting the voltage divider right(pch on the
high side would reverse the resistors and the cap would be to vcc rather
than to ground).

DaveC wrote in
:

Relay-logic controlled printing machine, 110 vac.

A 300mfd (150v) electrolytic cap is used together with 25K (25 W) ohm
wirewound pot and a 5K ohm relay coil to control a variable time delay
before the relay opens.

The cap is shot (measures less than 1mfd).

What was the existing cap? Any clear indications as to why or how it failed?

All you need is a resistor(actually two), a cap, a suitable mosfet that
can
handle the current and voltage(relatively easy to find for a few dollars),
and a diode for a snubber.

BTW, if ac is on the cap and not rectified you'll either want to add
rectification or go with a triac/scr version instead. Mosfets only work well
one way. I'm assuming the relay is DC since you mentioned an electrolytic
cap.

DaveC wrote in
:

Or a motor-start cap? ::

http://tinyurl.com/y9d3ks6


Another thought, if you ever need a high quality cap with high voltage and a
few hundred µF at short notice and can't buy one, look for disposable or
otherwise surplus camera flashes. they're usually good for electrolytics
at 350V at 200µF or more.

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?

A motor start cap should be ok. They're meant for intermittent use under
heavy load so they don't accept heavy ripple currents for lomg, but this one
won't have to, with a few tens of kilohms between it and the mains.

A single rectifier in series with 50 ohms feeds the charge current to the
cap. Seems that 150v is a bit overkill, no? (The 25K pot is on the discharge
side of the circuit which is switched in at the appropriate time.)

I think I'm going to go with the 270mfd / 125v start cap. It's sourceable
locally, fits the space, and although a bit low on the capacitance it will
fit the bill better than what's there.

DaveC wrote in
:

A motor start cap should be ok. They're meant for intermittent use
under heavy load so they don't accept heavy ripple currents for lomg,
but this one won't have to, with a few tens of kilohms between it and
the mains.

A single rectifier in series with 50 ohms feeds the charge current to
the cap. Seems that 150v is a bit overkill, no? (The 25K pot is on the
discharge side of the circuit which is switched in at the appropriate
time.)

I think I'm going to go with the 270mfd / 125v start cap. It's
sourceable locally, fits the space, and although a bit low on the
capacitance it will fit the bill better than what's there.



Don't worry if the cap is specced for a higher voltage, if it's cheap, have
at it, the margin of safety is good, the wider, the better.

DaveC wrote in
:

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. Heat could well be right. Also accelerated loss
of electrolyte. Did the cap look like it had burst from inside earlier than
any final appearance of internal gubbins? Another possibility is the voltage
it saw. When you say 110 volts, you mean the mains, right? If so it will be
rectified and that cap will be seeing the peak value, not the RMS, so around
125V when under no load, so you really do need it to have a higher voltage
rating than that.

Lostgallifreyan wrote in
:

DaveC wrote in
:

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. Heat could well be right. Also accelerated
loss of electrolyte. Did the cap look like it had burst from inside
earlier than any final appearance of internal gubbins? Another
possibility is the voltage it saw. When you say 110 volts, you mean the
mains, right? If so it will be rectified and that cap will be seeing the
peak value, not the RMS, so around 125V when under no load, so you
really do need it to have a higher voltage rating than that.

Btw, I know that that cap ought not to be unloaded to the point where the
voltage gets to full height in that system, but best assume it will, so if
the load fails the cap remains safe and you get no bad future comeback. I've
seen an amplifier that had an internal fire resulting from a cap failure. It
shocked me. Very very messy scene. I wasn't given to me for repair, I didn't
even get much salvage out of it.

On Wed, 07 Oct 2009 22:27:31 -0500, Lostgallifreyan
wrote:

DaveC wrote in
al-september.org:

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. Heat could well be right. Also accelerated loss
of electrolyte. Did the cap look like it had burst from inside earlier than
any final appearance of internal gubbins? Another possibility is the voltage
it saw. When you say 110 volts, you mean the mains, right? If so it will be
rectified and that cap will be seeing the peak value, not the RMS, so around
125V when under no load, so you really do need it to have a higher voltage
rating than that.

---

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

John Fields wrote in
:

On Wed, 07 Oct 2009 22:27:31 -0500, Lostgallifreyan
wrote:

DaveC wrote in
l-september.org:

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. Heat could well be right. Also accelerated
loss of electrolyte. Did the cap look like it had burst from inside
earlier than any final appearance of internal gubbins? Another
possibility is the voltage it saw. When you say 110 volts, you mean the
mains, right? If so it will be rectified and that cap will be seeing the
peak value, not the RMS, so around 125V when under no load, so you
really do need it to have a higher voltage rating than that.

---

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF


Agreed. I even knew the formula and entered exactly 1.414, no less, so I'm
still wondering what the hell else I wrote into the Windows calculator this
morning to get an answer of fractionally over 125V. I'll put this down as yet
another reason to aim for safe margins if the cost doesn't prevent it.

(Actually I found it. I typoed as 1.141...)

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

Thanks.

DaveC wrote in
:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

Thanks.



Yes, under no load either peak of a sine wave defines the maximum (absolute)
value for voltage across a capacitor. Full wave rectification just means more
energy and less ripple present for a given load.

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

Thanks.

Half-wave and full-wave rectification result in the same peak voltage.
It's just that for full-wave, the peaks occur twice as often.

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

My line voltage runs well above 110.

PG&E says they target for a min of 114 and a max of 126.
That's at the handoff point. I think the idea is that
you get a min of 110 if you allocate a few volts for
drop on the wiring within your house.


Here is PG&E's blurb:
http://www.pge.com/tariffs/tm2/pdf/ELEC_RULES_2.pdf


--
These are my opinions, not necessarily my employer's. I hate spam.

Yep, yep.. half-wave is a single rectifier, full-wave is 2 rectifiers often
with a center-tapped AC source, and full-wave-bridge rectification is 4
rectifiers.

--
Cheers,
WB
..............


"Mike Paff" wrote in message
...
On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

Thanks.

Half-wave and full-wave rectification result in the same peak voltage.
It's just that for full-wave, the peaks occur twice as often.

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.

John Fields wrote in
:

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.



So what IS the story, exactly? Right now you've focussed on the results of a
load (the resistor), so you've said things that might be taken to discredit
the rest of us who specified that either peak has an absolute value of
voltage across a capacitor that when unloaded, must be considered for its
safety. While a half-wave rectified form is a more complex wave whose RMS
value needs a different calculation, in this case it's the separation between
peaks that matters.

Many PSU capacitor failures seems to be a result of people underspecifying
the working voltage while assuming they'll be 'safely' loaded to prevent the
peak from dictating terms more than the RMS. It's not a safe assumption.

On Thu, 08 Oct 2009 22:27:38 -0500, Lostgallifreyan
wrote:

John Fields wrote in
:

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.



So what IS the story, exactly? Right now you've focussed on the results of a
load (the resistor), so you've said things that might be taken to discredit
the rest of us who specified that either peak has an absolute value of
voltage across a capacitor that when unloaded, must be considered for its
safety.

While a half-wave rectified form is a more complex wave whose RMS
value needs a different calculation, in this case it's the separation between
peaks that matters.

Many PSU capacitor failures seems to be a result of people underspecifying
the working voltage while assuming they'll be 'safely' loaded to prevent the
peak from dictating terms more than the RMS. It's not a safe assumption.

---
Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.

Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.

No worries, mate! I'm learned(er).

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.

Wow. That's probably why the original 150v cap is no good after 40 years. It
should have been spec'd higher. Is 160 good enough for this application?

Thanks.

DaveC wrote in
:

Is 160 good enough for this application?


Doubtful, note the post from Hal Murray, about how much line voltage can
vary. (Even at exactly 110V which it likely never is for long, you're pushing
to to within 5V of a dangerous limit and trusting capacitor voltage tolerance
is a Very Bad Idea, unless you're going to stress test samples in multiple
copies of a design).

You might get by for a while if the cap is run as cool as possible, but bear
in mind that the last one failed, and you now have a very good reason to
suspect it failed from being run too close to maximum levels. So leave a
generous margin this time.

On Fri, 9 Oct 2009 20:13:53 -0700, DaveC wrote:

Sorry for the confusion, and you're right, an unloaded cap will charge
to 1.414 times the RMS value of the unrectified sine wave, no matter
whether the output of the rectifier is full-wave or half-wave.

No worries, mate! I'm learned(er).

Chances are that a loaded one will also, since most power supplies are
designed to have the cap charge fully and then discharge, between peaks,
to yield an acceptable ripple.

Wow. That's probably why the original 150v cap is no good after 40 years. It
should have been spec'd higher. Is 160 good enough for this application?

---
Dunno...

Can you post a schematic?

Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing
Frank

"Lostgallifreyan" wrote in message
. ..
DaveC wrote in
:

What was the existing cap?

DaveC wrote in
:
It's a Mallory FP119A.

Any clear indications as to why or how it failed?

No. Guesses: Old age? Heat?



Old age isn't a mechanism. Heat could well be right. Also accelerated
loss
of electrolyte. Did the cap look like it had burst from inside earlier
than
any final appearance of internal gubbins? Another possibility is the
voltage
it saw. When you say 110 volts, you mean the mains, right? If so it will
be
rectified and that cap will be seeing the peak value, not the RMS, so
around
125V when under no load, so you really do need it to have a higher voltage
rating than that.

"Frank S" wrote in
:

Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing


I agree. Was just saying that length of time is like the length of a piece of
string. Like someone else here said, some electrolytic caps work fine in gear
tens of years old. Valve/tube gear even, where you have enough heat to
accerate drying. It sort of suggests that looking at lengths of time is
missing plenty, there's a lot of difference between electrolytics, they're
not at all consistent.

"Lostgallifreyan" wrote in message
. ..
"Frank S" wrote in
:

Old age isn't a mechanism, but is a cause for caps to go bad old age =
drying, drying = dieing


I agree. Was just saying that length of time is like the length of a piece
of
string. Like someone else here said, some electrolytic caps work fine in
gear
tens of years old. Valve/tube gear even, where you have enough heat to
accerate drying. It sort of suggests that looking at lengths of time is
missing plenty, there's a lot of difference between electrolytics, they're
not at all consistent.

And there's how hard the user pushes them as regards the voltage and
temperature
ratings.